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The equivalent conductivity of `KCl` at infinite dilution is `130S cm^(2) eq^(-1)`. The transport number of `Cl^(-)` ion in `KCl` at the same temperature is `0.505`. The limiting ionic mobility of `K^(+)` ion is :A. `6xx67xx10^(-4)cm^(2)sec^(-1)"volt"^(-1)`B. `5.01xx10^(-3)cm^(2)sec^(-1)"volt"^(-1)`C. `3.22xx10^(-4)cm^(2)sec^(-1)"volt"^(-1)`D. `2.00xx10^(-4)cm^(2)sec^(-1)"volt"^(-1)` |
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Answer» Correct Answer - A `(lamda_(cl^(-))^(@))/(lamda_(K^(+))^(@)+lamda_(Cl^(-))^(@))0.505` or `lamda_(Cl^(-))^(@)=0.505xx130approx65.65Scm^(2)eq^(-1)` `lamda_(K^(+))^(@)=FxxU_(K^(+))` or `U_(K^(+))=((130-65.65))/(96500)cm^(2)"volt"^(-1)sec^(-1)` |
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